Question

P=\(\dfrac{3\sqrt{x}+6}{x-4}\) -\(\dfrac{\sqrt{x}}{2-\sqrt{x}}\)

Rút gọn

Answer 1

P = \(\dfrac{3\sqrt{x}+6}{x-4}-\dfrac{\sqrt{x}}{2-\sqrt{x}}\) (x \(\ne\) 4; x \(\ge\) 0)

P = \(\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\)

P = \(\dfrac{3\sqrt{x}+6+\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\)

P = \(\dfrac{3\sqrt{x}+6+x+2\sqrt{x}}{x-4}\)

P = \(\dfrac{x+5\sqrt{x}+6}{x-4}\)

P = \(\dfrac{x+2\sqrt{x}+3\sqrt{x}+6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

P = \(\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)+3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

P = \(\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

P = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)

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Answer 2

Lời giải:

ĐKXĐ: $x\neq 4; x\geq 0$

\(P=\frac{3(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{3}{\sqrt{x}-2}+\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{3+\sqrt{x}}{\sqrt{x}-2}\)

Answer 3

Ta có: \(\dfrac{3\sqrt{x}+6}{x-4}-\dfrac{\sqrt{x}}{2-\sqrt{x}}\)

\(=\dfrac{3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{3}{\sqrt{x}-2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

\(=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)